Question: Evaluate $~~\int^e_1 x^3\ln x\ dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{3e^4}{16}+\dfrac1{16}$ (Choice B) B $\dfrac{5e^4}{16}-\dfrac1{16}$ (Choice C) C $\dfrac{2e^4}{9}+\dfrac1{9}$ (Choice D) D $\dfrac{2e^3}{9}+\dfrac1{9}$
We will solve this by integrating by parts. We know that $ \int u(x)v\,^\prime(x)dx = u(x)v(x)-\int u\,^\prime(x)v(x)dx\,$. We can rewrite this as $ \int u\ dv = uv-\int v\ du\,$. In this problem we will let $~u = \ln x~$ and $~dv=x^3\ dx\,$. Then $~du = \dfrac1xdx~$ and $~v = \int x^3dx = \dfrac14x^4\,$. Integration by parts gives $ \int^e_1 x^3\ln x\,dx = \ln x\cdot\dfrac14x^4\Bigg]_1^e-\int^e_1\dfrac14x^4\cdot\dfrac1xdx$ $ ~~~~~~=\dfrac{x^4\ln x}4\Bigg]_1^e-\int^e_1\dfrac{x^3}4dx$ $ ~~~~~~=\dfrac{x^4\ln x}4-\dfrac1{16}x^4\Bigg]^e_1$ $ ~~~~~~=\Bigg(\dfrac {e^4}4-\dfrac{e^4}{16}\Bigg)-\Bigg(-\dfrac1{16}\Bigg) = \dfrac{3e^4}{16}+\dfrac1{16}\,$.